Question

A capacitor of capacitance C connected to a ideal DC battery of emf 40 V through a resistance R at t = 0. It is found that potential difference across the capacitor increases by 10 V in 2 $\mu$ sec.

The time after which the charge on capacitor became 50% of its maximum value is $\frac{ln(4)}{ln(x)-ln(y)}\mu$ sec. Find (x - y), where x and y are co-prime integers.

Answer 1

1

Answer 2

The problem describes the charging of a capacitor in an RC circuit. The voltage across a charging capacitor at time $t$ is given by:

$V_C(t) = V_0 (1 - e^{-t/RC})$

where $V_0$ is the battery emf and $RC$ is the time constant ($\tau$).

1. Determine the time constant (RC):
   Given $V_0 = 40$ V.
   At $t = 2 \, \mu$sec, the potential difference across the capacitor $V_C(2 \, \mu\text{sec})$ increases by 10 V, so $V_C(2 \, \mu\text{sec}) = 10$ V (assuming it was uncharged at $t=0$).
   Substitute these values into the equation:

   $10 = 40 (1 - e^{-2\, \mu\text{sec}/RC})$ $\frac{10}{40} = 1 - e^{-2\, \mu\text{sec}/RC}$ $\frac{1}{4} = 1 - e^{-2\, \mu\text{sec}/RC}$ $e^{-2\, \mu\text{sec}/RC} = 1 - \frac{1}{4}$ $e^{-2\, \mu\text{sec}/RC} = \frac{3}{4}$ Take the natural logarithm of both sides:

   $-\frac{2\, \mu\text{sec}}{RC} = \ln\left(\frac{3}{4}\right)$ $-\frac{2\, \mu\text{sec}}{RC} = \ln(3) - \ln(4)$ $\frac{2\, \mu\text{sec}}{RC} = \ln(4) - \ln(3)$ From this, the time constant $RC$ is:

   $RC = \frac{2\, \mu\text{sec}}{\ln(4) - \ln(3)}$

2. Find the time when the charge on the capacitor becomes 50% of its maximum value:
   The charge on the capacitor at time $t$ is given by $Q(t) = Q_{max} (1 - e^{-t/RC})$, where $Q_{max} = C V_0$.
   We want to find the time $t'$ when $Q(t') = 0.5 Q_{max}$.

   $Q_{max} (1 - e^{-t'/RC}) = 0.5 Q_{max}$ $1 - e^{-t'/RC} = 0.5$ $e^{-t'/RC} = 0.5$ $e^{-t'/RC} = \frac{1}{2}$ Take the natural logarithm of both sides:

   $-\frac{t'}{RC} = \ln\left(\frac{1}{2}\right)$ $-\frac{t'}{RC} = -\ln(2)$ $\frac{t'}{RC} = \ln(2)$ $t' = RC \ln(2)$

3. Substitute the value of RC:
   Substitute the expression for $RC$ from step 1 into the equation for $t'$:

   $t' = \left( \frac{2\, \mu\text{sec}}{\ln(4) - \ln(3)} \right) \ln(2)$ $t' = \frac{2 \ln(2)}{\ln(4) - \ln(3)}\, \mu\text{sec}$ Using the logarithm property $n \ln(a) = \ln(a^n)$, we have $2 \ln(2) = \ln(2^2) = \ln(4)$.

   $t' = \frac{\ln(4)}{\ln(4) - \ln(3)}\, \mu\text{sec}$

4. Compare with the given format and find (x - y):
   The given format for the time is $\frac{\ln(4)}{\ln(x)-\ln(y)}\mu$ sec.
   By comparing our derived expression for $t'$ with the given format, we can identify $x$ and $y$:
   $x = 4$
   $y = 3$
   The problem states that $x$ and $y$ are co-prime integers, which 4 and 3 are (their greatest common divisor is 1).
   Finally, calculate $(x - y)$:

   $x - y = 4 - 3 = 1$

The final answer is $\boxed{1}$.

Explanation of the solution:

1. The voltage across a charging capacitor is $V_C(t) = V_0 (1 - e^{-t/RC})$.
2. Given $V_0 = 40$ V and $V_C(2 \, \mu\text{sec}) = 10$ V. Substitute these to find the time constant $RC$:
   $10 = 40 (1 - e^{-2\, \mu\text{sec}/RC}) \implies e^{-2\, \mu\text{sec}/RC} = 3/4$.
   Taking natural log: $-\frac{2\, \mu\text{sec}}{RC} = \ln(3/4) = \ln(3) - \ln(4)$.
   So, $RC = \frac{2\, \mu\text{sec}}{\ln(4) - \ln(3)}$.
3. The charge on the capacitor is $Q(t) = Q_{max} (1 - e^{-t/RC})$. For the charge to be 50% of its maximum value, $Q(t') = 0.5 Q_{max}$. This implies $1 - e^{-t'/RC} = 0.5$, so $e^{-t'/RC} = 0.5 = 1/2$.
4. Taking natural log: $-\frac{t'}{RC} = \ln(1/2) = -\ln(2)$. So, $t' = RC \ln(2)$.
5. Substitute the expression for $RC$ into $t'$:
   $t' = \left( \frac{2\, \mu\text{sec}}{\ln(4) - \ln(3)} \right) \ln(2) = \frac{2 \ln(2)}{\ln(4) - \ln(3)}\, \mu\text{sec}$.
6. Using the property $2 \ln(2) = \ln(2^2) = \ln(4)$:
   $t' = \frac{\ln(4)}{\ln(4) - \ln(3)}\, \mu\text{sec}$.
7. Comparing this with the given format $\frac{\ln(4)}{\ln(x)-\ln(y)}\mu$ sec, we find $x=4$ and $y=3$. These are co-prime integers.
8. Calculate $(x - y) = 4 - 3 = 1$.